If a machine consumes 5 kilowatts of power every t hours, how much pow

Rate and algebra
The fractional number of kilowatts in the prompt might obscure the obvious: "___ kilowatts every ___ hours" is a rate.

What is the rate of consumption for one machine?
Number of kilowatts used: k/5
Per time unit: every t hours

Rate of consumption (no units after first term):
\(\frac{\frac{k}{5}kws}{t_{hours}}=(\frac{k}{5}*\frac{1}{t})=\frac{k}{5t}\)

At that rate, how much power will three such machines consume in 10 hours?

(Consumption rate) * (time spent consuming, 10 hrs) * (# of consumers, 3) = Total consumption
= Amount of power used

\((\frac{k}{5t} * 10 * 3) = \frac{30k}{5t}=\frac{6k}{t}\) kilowatts

Answer E

Assign values
Assign values that result in integers.
Let k = 5
Let t = 1

Rate of power consumption
One machine consumes k/5 kilowatts of power every t hours

\(\frac{k}{5}kw=\frac{5}{5}kw = 1\) kilowatt is consumed . . .
Every \(t = 1\) hours
Consumption rate is \(\frac{1kw}{1hr}\)

How much power, in kilowatts, will three such machines consume in 10 hours?

(Consumption rate) * (# of machines) * (Time) = Total consumption

\(\frac{1kilowatt}{1hr} * 3 * 10hrs = 30\) kilowatts

With k = 5, t = 1, find the answer choice that yields 30.

At a glance with mental math, eliminate B and D immediately (much too small), and C (huge). That leaves A and E

A. 6t/k: \(\frac{(6*1)}{5} = \frac{6}{5}\). NO

E. 6k/t: \(\frac{(6*5)}{1} = \frac{30}{1} = 30\). MATCH

Answer E

*The eliminated options, which are indeed small or huge compared to 30:
B. t/k: \(\frac{1}{5}\). NO

C. 30kt: \((30)*(5)*(1) = 150\). NO

D. k/t: \(\frac{5}{1} = 5\). NO